Prove that x = - b±√b²-4ac / 29

Using logarithm properties $\log(MN) = \log M + \log N$ and $\log(M^p) = p \log M$:

$$\log_{10} 12 = \log_{10}(3 \times 4) = \log_{10}(3 \times 2^2)$$ $$\log_{10} 12 = \log_{10} 3 + 2 \log_{10} 2$$

Substitute the given values:

$$\log_{10} 12 = 0.4771 + 2(0.3010)$$ $$\log_{10} 12 = 0.4771 + 0.6020$$ $$\log_{10} 12 = 1.0791$$

Using logarithm property $\log(M/N) = \log M - \log N$:

$$\log_{10}\left(\frac{3}{2}\right) = \log_{10} 3 - \log_{10} 2$$

Substitute the given values:

$$\log_{10}\left(\frac{3}{2}\right) = 0.4771 - 0.3010$$ $$\log_{10}\left(\frac{3}{2}\right) = 0.1761$$

$$\frac{\log_{10} 12}{\log_{10}(3/2)} = \frac{1.0791}{0.1761}$$ $$\frac{1.0791}{0.1761} \approx 6.127768...$$

Rounding to four decimal places:

$$\frac{\log_{10} 12}{\log_{10}(3/2)} \approx \boxed{6.1278}$$

$$ax^2 + bx + c = 0$$

Assuming $a \neq 0$, divide the entire equation by $a$:

$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$

Add $\left(\frac{b}{2a}\right)^2$ to both sides:

$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$$

Rewrite the left side as a perfect square:

$$\left(x + \frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$$

Find a common denominator ($4a^2$) for the terms on the right side:

$$\left(x + \frac{b}{2a}\right)^2 = -\frac{4ac}{4a^2} + \frac{b^2}{4a^2}$$ $$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$

$$x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$$

Simplify the square root:

$$x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$

Subtract $\frac{b}{2a}$ from both sides:

$$x = -\frac{b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$

Combine the terms since they have a common denominator:

$$x = \boxed*{\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$$